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3 years ago

PLEASE HELP I CANT CONTINUE WITH MY WORK UNLESS THESE ARE COMPLETE AND CORRECT

Mathematics

1 answer:

Snezhnost [94]3 years ago

6 0

Answer:

1) x=3, y=-3

Step-by-step explanation:

10x + 7y = 9 —— (1)

-4x -7y = 9 —— (2)

(1) + (2)

6x = 18

x = 18/6

x = 3

put x=3 in (1)

10(3) + 7y = 9

30 + 7y = 9

7y = 9 - 30

7y = -21

y = -21/7

y = -3

please give me brainlist if it is helpful for you

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Given:

The rate of interest on three accounts are 7%, 8%, 9%.

She has twice as much money invested at 8% as she does in 7%.

She has three times as much at 9% as she has at 7%.

Total interest for the year is $150.

To find:

Amount invested on each rate.

Solution:

Suppose x be the amount invested at 7%. Then,

The amount invested at 8% = 2x

The amount invested at 9% = 3x

Total interest for the year is $150.

$x\times \dfrac{7}{100}+2x\times \dfrac{8}{100}+3x\times \dfrac{9}{100}=150$

Multiplying both sides by 100, we get

$7x+16x+27x=15000$

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Dividing both sides by 50, we get

$x=\dfrac{15000}{50}$

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So, the amount invested at 7% is $300$ .

The amount invested at 8% is

$2(300)=600$

The amount invested at 9% is

$3(300)=900$

Hence, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.

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